Question: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 26,\enspace 10,\enspace 8,\enspace 2,\enspace 12$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{26 + 10 + 8 + 2 + 12}{{5}} = {11.6\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $26$ years $14.4$ years $207.36$ years $^2$ $10$ years $-1.6$ years $2.56$ years $^2$ $8$ years $-3.6$ years $12.96$ years $^2$ $2$ years $-9.6$ years $92.16$ years $^2$ $12$ years $0.4$ years $0.16$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{207.36} + {2.56} + {12.96} + {92.16} + {0.16}} {{5}} $ $ {\sigma^2} = \dfrac{{315.2}}{{5}} = {63.04\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{63.04\text{ years}^2}} = {7.9\text{ years}} $ The average gorilla at the zoo is 11.6 years old. There is a standard deviation of 7.9 years.